Find the points on the surface $z ^ { 2 } = x y + 49$ that are closest to the origin.
Select the correct answer.
A) $( 0,0 , - 49 )$
B) $( 0,0,49 ) ( 0,0 , - 49 )$
C) $( 0,7,0 )$
D) $( 0,0,7 ) ( 0,0 , - 7 )$
E) $( 0,0,7 )$

Question

Quizplus · Accepted Answer

We want to minimize the distance from the origin, which is given by $\sqrt{x^2+y^2+z^2}$. This is equivalent to minimizing $x^2+y^2+z^2$. By inspection, we see that it's possible for $z$ to be equal to $0$. Thus, we can minimize $x^2+y^2$ subject to the constraint $z^2=xy+49$. This can be done using Lagrange multipliers, but it's easier to use a bit of insight.

One possible way is to note that the constraint says $xy+49\geq0$, which implies that $z^2\geq0\implies x^2+y^2\geq49$. This suggests that the minimum is achieved on the boundary of the circle $x^2+y^2=49$, where $xy+49=49$ is a line. Then $z^2=0$, which means $z=0$, and $x^2+y^2=49$ implies that the closest points are $(0,0,\pm7)$, which are on the boundary of the circle. Thus, the answer is $\boxed{	extbf{(D)}}$.

Find the Points on the Surface z2=xy+49z ^ { 2 } = x y + 49z2=xy+49

Find the Points on the Surface $z ^ { 2 } = x y + 49$